What is the acceleration due to gravity at a distance of 2 earth radii above earth's surface?

Question

The exact question which I'm having trouble with:

What is the acceleration due to gravity at a distance of 2 Earth radii above Earth's surface?

I have, no idea what this thing is asking. Isn't a radius a distance between the center of an object and its outer shell? So is this thing asking me the pull of gravity on an object, by earth, if the object is twice the distance from earth's surface than the surface and the core? What is the equation for that?

Answer 1

Newton's law of universal gravitation is
F = G*m1*m2/r^2
That tells you the force of attraction between any 2 objects (m1 and m2) at a separation distance of r. Since
F = m*a
if m1 is an object on, or near, the earth's surface, the acceleration that an object of mass m1 would experience if you dropped it from a meter or 2 above ground could be found as follows:
F = m1*a = G*m1*m2/r^2
a = G*m2/r^2
So that acceleration a is what we call g, 9.8 m/s^2.

In the above, r = the earth's radius. Your question asks what is the acceleration if r is replaced by 3*r (2 radii above the surface is 3*r from the center). Plug that in for r and evaluate. See what the difference is. (You don't need to know G, r, or m2 to compare the 2 results.)

Answer 2

enable earth radius be R,then entire distance of factor of remark from middle of earth=3R enable 'g' be the fee of acceleration by way of gravity on the exterior g=GM/R^2 the place M is mass of earth enable gh be the fee at wanted factor gh=GM/(3R)^2=GM/9(R^2) gh / g = { R / 3R} ^2= a million /9 Acceleration by way of gravity at top 2R above the exterior i.e. gh would be g / 9